The second-order reaction 2 Mn(CO)5 → Mn2(CO)10, has a rate constant equal to 3.0 × 109 M-1 s-1 at 25°C.?
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The second-order reaction 2 Mn(CO)5 → Mn2(CO)10, has a rate constant equal to 3.0 × 109 M-1 s-1 at 25°C. If the initial concentration of Mn(CO)5 is 3.0 × 10-5 M, how long will it take for 90.% of the reactant to disappear?
I'd appreciate the help.
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Use integrated second order rate law, which has the general form [1]:
(1/[A]) = k∙t + (1/[A]₀)
(with [A] reactant concentration at time t, k rate constant, [A]₀ initial reactant concentration i.e. at t=0) For your reaction A is manganese pentacarbonyl Mn(CO)₅.
Let X be the fraction of reactant which is consumed at time t, i.e.
X = -Δ[A] / [A]₀ = ([A]₀ - [A] ) / [A]₀ = 1 - ( [A]/[A]₀ )
<=>
[A] = [A]₀∙(1 - X)
Hence,
(1/( [A]₀∙(1 - X) ) ) = k∙t + (1/[A]₀)
<=>
1/(1 - X) = [A]₀∙k∙t + 1
When solve this relation for t you get the time elapsed until a specified degree of conversion is reached:
t = (1/(1 - X)) - 1) / ([A]₀∙k)
= X / ( [A]₀∙k∙(1 - X) )
For this problem
t = X / ( [Mn(CO)₅]₀∙k∙(1 - X) )
= 0.9 / ( 3.0×10⁻⁵ M ∙ 3.0×10⁹ M⁻¹∙s⁻¹ ∙(1 - 0.9) )
= 1.0×10⁻⁴ s
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